POJ 3767 I Wanna Go Home

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略水。。

题目大意是有n个城市，要从第1个城市走到第2个城市的最短路。。(起点是1，终点是2)，但是，每个城市都属于不同的领主。。

"For the sake of safety,", said Mr.M, "your route should contain at most 1 road which connects two cities of different camp.

要求你走的最短路中，最多包含1条连接两个城市但却属于不同的领主的路。。

dijkstra算法，+多一个sum[]数组记录有多少条连接两个城市但却属于不同的领主的路。。

#include<iostream>#include<cstring>#include<cstdio>#include<string>#include<algorithm>#include<cmath>#include<queue>#include<vector>using namespace std;#define eps 1e-8#define mod 1000000007#define mnx 20005#define ll long long #define ull unsigned long long#define inf 0x3f3f3f3fint cost[mnx], first[mnx], vv[mnx], nxt[mnx], dis[mnx], lead[mnx], sum[mnx], e, m, n;bool vis[mnx];struct edge{int u, d;bool operator < (const edge &b) const {return d > b.d;}};void init(){memset(dis, 0x3f, sizeof(dis));memset(vis, 0, sizeof(vis));memset(first, -1, sizeof(first));memset(sum, 0, sizeof(sum));e = 0;}void add( int u, int v, int c ){cost[e] = c;vv[e] = v;nxt[e] = first[u];first[u] = e++;}void dijkstra(int s){priority_queue<edge> que;dis[s] = 0;edge q1; q1.u = s; q1.d = 0;que.push(q1);while( !que.empty() ){q1 = que.top(); que.pop();int u = q1.u;if( vis[u] ) continue;vis[u] = 1;for( int i = first[u]; i != -1; i = nxt[i] ){int v = vv[i];if( dis[v] > dis[u] + cost[i] ){if(lead[u] == lead[v] || (lead[u] != lead[v] && sum[u] + 1 < 2) ) {  //注意条件。。     if( lead[u] != lead[v] ) sum[v] = sum[u] + 1;    else sum[v] = sum[u];    dis[v] = dis[u] + cost[i];    edge q2;    q2.d = dis[v], q2.u = v;    que.push(q2);}}}}}int main(){while( scanf("%d", &n) && n ){init();scanf("%d", &m);int u, v, d;for( int i = 0; i < m; i++ ){scanf("%d%d%d", &u, &v, &d);add(u, v, d);add(v, u, d);}for( int i = 1; i <= n; i++ ){scanf("%d", &lead[i]);}dijkstra(1);if( dis[2] >= inf ) printf("-1\n");else printf("%d\n", dis[2]);}return 0;}

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